Two charges -Q o and -3Q are sdistance l apart.These two charges are free to mov

Question

Two charges -Qo and -3Q are sdistance l apart.These two charges are free to move but do not because there is athird charge nearby. What must be the the charge and placement ofthe third charge for the irst two to be in equilibrium?  
I have noticed that somehow the x value is .366l and would like a break down of how that actually happened as well... thx... Two charges -Qo and -3Q are sdistance l apart.These two charges are free to move but do not because there is athird charge nearby. What must be the the charge and placement ofthe third charge for the irst two to be in equilibrium?  
I have noticed that somehow the x value is .366l and would like a break down of how that actually happened as well... thx...

Explanation / Answer

that means net force on both charges is zero

so

k 3 Q0^2/l^2 = k Q *q/x^2

simpligy

3 Q/l^2 = q/x^2

q = 3 Q x^2/l^2

and

k 3 Q0^2/l^2 = k 3 Q *q/(l-x)^2

simpligy

Q/l^2 = q/(l-x)^2

plug in for q

Q/l^2 = 3 Q x^2/l^2/(l-x)^2

(l-x)^2 = 3 x^2

l - x = sqrt(3) x

x = 1/(sqrt(3) + 1) l=0.366 l

q = 3*0.366^2 Q=0.402 Q

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